How to prove this algebraic version of the sine law?

Question

How to solve the following problem from Hall and Knight's Higher Algebra?

Suppose that \begin{align} a&=zb+yc,\tag{1}\\ b&=xc+za,\tag{2}\\ c&=ya+xb.\tag{3} \end{align} Prove that $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$

(I suppose that $x,y,z$ are real numbers whose moduli are not equal to $1$.)

I discovered this problem from chapter 3 of Prelude to Mathematics by W. W. Sawyer. Sawyer thought that this problem arose from the sine law: let $a,b,c$ be respectively the lengths of the edges opposite to three vertices $A,B,C$ of a triangle. Define $x=\cos A$ and define $y,z$ analogously. Now equalities $(1)-(3)$ simply relate $a,b$ and $c$ to each other by the cosines of the angles and $(4)$ is just a rewrite of the sine law $$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. $$ However, the algebraic version $(4)$ looks more general. For example, it does not state that $a,b,c$ must be positive or that they must satisfy the triangle inequality.

Sawyer wrote that this isn't a hard problem, but he didn't provide any solution. I can prove $(4)$ using linear algebra. Suppose that $(a,b,c)\ne(0,0,0)$ (otherwise $(4)$ is obvious). Rewrite $(1)-(3)$ in the form of $M\mathbf a=0$: $$\begin{bmatrix}-1&z&y\\ z&-1&x\\ y&x&-1\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=0.$$ Since $x^2,y^2,z^2\ne1$, $M$ has rank $2$ and $D=\operatorname{adj}(M)$ has rank $1$. Hence all columns of $D$ are parallel to $(a,b,c)^T$ and $\frac{d_{11}}{d_{21}}=\frac{d_{12}}{d_{22}}=\frac{a}{b}$. Since $M$ is symmetric, $D$ is symmetric too. Therefore $\frac{1-x^2}{1-y^2}=\frac{d_{11}}{d_{22}}=\frac{d_{11}d_{12}}{d_{21}d_{22}}=\frac{a^2}{b^2}$, i.e. $\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.

As this problem comes from Hall and Knight's book, I think there should be a more elementary solution. Any ideas?

Answer 1

Let $a=0$.

Thus, $$xc=b$$ and $$xb=c,$$ which gives $$x^2bc=bc$$ or $$(x^2-1)bc=0$$ and since $x^2\neq1,$ we obtain $bc=0$ and from here $$a=b=c=0,$$ which gives that our statement is true.

Let $abc\neq0$.

Thus, $$\frac{zb}{a}+\frac{yc}{a}=1$$ and $$\frac{xc}{b}+\frac{za}{b}=1,$$ which gives $$z^2+\frac{xyc^2}{ab}+\frac{xzc}{a}+\frac{yzc}{b}=1$$ or $$\frac{1-z^2}{c^2}=\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca},$$ which gives $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}=\frac{1}{\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}}.$$

Answer 2

We can write

I) $z=\frac{a-yc}{b}$ from (1)

Now, multiplying $y$ on both sides of (3) we get $yc=y^2a+xyb$.

So, from I) we get $\frac{a-ay^2}{b}-xy=z$.....(1')

Similarly from equation (2) we get

II) $z=\frac{b-xc}{a}$ from (2)

Now, multiplying $x$ on both sides of (3) we get $xc=xya+x^2b$.

And we get from (2) $\frac{b-bx^2}{a}-xy=z$.....(2')

From (1') and (2') we get,

$\frac{a-ay^2}{b}=\frac{b-bx^2}{a} \rightarrow \frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.

Similarly, $\frac{a^2}{1-x^2}=\frac{c^2}{1-z^2}$

So, ultimately we have proved $\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}$

Suppose, $b=0$, then $\frac{a}{c}=y=\frac{c}{a}$ or $y=1$. So, $\frac{b^2}{1-y^2}$ will be undefined.

If also $c=0$ then $a=0$.

Answer 3

It turns out that I solved the equations for the wrong variables. If I rewrite $(1)-(3)$ as $$\begin{bmatrix}0&c&b\\ c&0&a\\ b&a&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}a\\ b\\ c\end{bmatrix}$$ and solve for $x,y,z$ instead, I will get the law of cosines, i.e. $$x=\frac{b^2+c^2-a^2}{2bc}$$ etc.. Therefore $$\frac{a^2}{1-x^2}=\frac{4a^2b^2c^2}{2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)}.$$ As Roman Odaisky has pointed out, this expression can be rewritten as $\frac{a^2b^2c^2}{4s(s-a)(s-b)(s-c)}$, where $s=\frac12(a+b+c)$. By symmetry, $\frac{b^2}{1-y^2}$ and $\frac{c^2}{1-z^2}$ are also equal to the same expression. Geometrically (and according to Heron's formula), this means the common ratio in the law of sines is equal to $\frac{abc}{2T}$ where $T$ is the area of the triangle.

Answer 4

By (1) and (3), $a=ay^2 + bxy +bz.$ Thus, $a(1-y^2)=b(xy+z)$ so that $$a^2(1-y^2)=ab(xy+z).$$ In a similar way, we derive from (2) and (3) that $$b^2(1-x^2)=ab(xy+z).$$ Thus, the left sides of the two displayed equations are equal, yielding the first equality in (4). By symmetry, we're done.

IOW replace $(a,c)$ by $(c,a)$ and $(x,z)$ by $(z,x)$ above.

Answer 5

$a(1)-b(2)$ gives $$a^2-b^2=c(ya-xb)$$ and with $(3)$ we get $$a^2-b^2=(ya+xb)(ya-xb)$$ $$a^2-b^2=y^2a^2-x^2b^2$$ $$a^2-a^2y^2=b^2-b^2x^2$$ $$a^2(1-y^2)=b^2(1-x^2)$$ $$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$$ cyclicly.